The last identity follows from the following identity which can be found in the proof of [1]}.
\(\frac{1}{2\pi }\int _{0}^{2\pi }K_0\big (\tfrac{2}{\alpha }\sin \big (\tfrac{\theta }{2}\big )\big )\cos (n\theta )d\theta =I_{n}\left(\tfrac{1}{\alpha }\right)K_{n}\left(\tfrac{1}{\alpha }\right).\) 
